Question

Light of wavelength 500 nm is incident perpendicularly from air on a film 10-4cm thick and of refractive index 1.375. Part of the light enters the film and is reflected back at the second surface.

a) How many "wavelengths" are contained along the path of this light in the film?
b) What is the phase difference between these reflected waves as they leave the film at the first surface?

Answer 1

Answer

given,

wavelength (λ)= 500 n m

thickness of film= 10⁻⁴ cm

refractive index = μ = 1.375

distance traveled is double which is equal to 2 x 10⁻⁴ cm

a) Number of wave

     $N = \dfrac{d}{\mu\lambda}$

     $N = \dfrac{2 \times 10^{-6}}{1.375\times 500 \times 10^{-9}}$

           N = 2.91

           N = 3

b) phase difference is equal to

Reflection from the first surface has a 180° (½λ) phase change.

There is no phase change for the 2nd surface reflection and there is no phase difference for the 2nd wave having traveled an exact whole number of waves.

net phase difference = $180^0\times \dfrac{3}{2}$

                                   = 270°